The F- will react with water to form HF and OH F-.pdf
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The F- will react with water to form HF and OH F- + H2O====> HF + OH- Kb for this reaction is 1.58 x 10^-11 At equilibrium, [F-] = .17 - x [HF] = x [OH-] = x The reactant is consumed by some amount x while the products are formed by some amount x Kb = concentration of products/reactants = x*x/(.17-x) = 1.58x 10^-11 x^2 + 1.58 x 10^-11x - 2.68 x 10^-12 = 0 Solve for x, x = 1.64 x 10^-6 [HF]= x = 1.64 x 10^-6 M [OH-] = x = 1.64 x 10^-6 M We know, [H3O+]*[OH-] = 10^-14 [H3O+] = 10^-14/(1.64 x 10^-6) = 6.09 x 10^-9 Hope this helped Solution The F- will react with water to form HF and OH F- + H2O====> HF + OH- Kb for this reaction is 1.58 x 10^-11 At equilibrium, [F-] = .17 - x [HF] = x [OH-] = x The reactant is consumed by some amount x while the products are formed by some amount x Kb = concentration of products/reactants = x*x/(.17-x) = 1.58x 10^-11 x^2 + 1.58 x 10^-11x - 2.68 x 10^-12 = 0 Solve for x, x = 1.64 x 10^-6 [HF]= x = 1.64 x 10^-6 M [OH-] = x = 1.64 x 10^-6 M We know, [H3O+]*[OH-] = 10^-14 [H3O+] = 10^-14/(1.64 x 10^-6) = 6.09 x 10^-9 Hope this helped.
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![The F- will react with water to form HF and OH F- + H2O====> HF + OH- Kb
for this reaction is 1.58 x 10^-11 At equilibrium, [F-] = .17 - x [HF] = x [OH-] = x The reactant
is consumed by some amount x while the products are formed by some amount x Kb =
concentration of products/reactants = x*x/(.17-x) = 1.58x 10^-11 x^2 + 1.58 x 10^-11x - 2.68 x
10^-12 = 0 Solve for x, x = 1.64 x 10^-6 [HF]= x = 1.64 x 10^-6 M [OH-] = x = 1.64 x 10^-6 M
We know, [H3O+]*[OH-] = 10^-14 [H3O+] = 10^-14/(1.64 x 10^-6) = 6.09 x 10^-9 Hope this
helped
Solution
The F- will react with water to form HF and OH F- + H2O====> HF + OH- Kb
for this reaction is 1.58 x 10^-11 At equilibrium, [F-] = .17 - x [HF] = x [OH-] = x The reactant
is consumed by some amount x while the products are formed by some amount x Kb =
concentration of products/reactants = x*x/(.17-x) = 1.58x 10^-11 x^2 + 1.58 x 10^-11x - 2.68 x
10^-12 = 0 Solve for x, x = 1.64 x 10^-6 [HF]= x = 1.64 x 10^-6 M [OH-] = x = 1.64 x 10^-6 M
We know, [H3O+]*[OH-] = 10^-14 [H3O+] = 10^-14/(1.64 x 10^-6) = 6.09 x 10^-9 Hope this
helped](https://image.slidesharecdn.com/thef-willreactwithwatertoformhfandohf-230411152857-bfbffb72/85/The-F-will-react-with-water-to-form-HF-and-OH-F-pdf-1-320.jpg)
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The F- will react with water to form HF and OH F-.pdf
- 1. The F- will react with water to form HF and OH F- + H2O====> HF + OH- Kb for this reaction is 1.58 x 10^-11 At equilibrium, [F-] = .17 - x [HF] = x [OH-] = x The reactant is consumed by some amount x while the products are formed by some amount x Kb = concentration of products/reactants = x*x/(.17-x) = 1.58x 10^-11 x^2 + 1.58 x 10^-11x - 2.68 x 10^-12 = 0 Solve for x, x = 1.64 x 10^-6 [HF]= x = 1.64 x 10^-6 M [OH-] = x = 1.64 x 10^-6 M We know, [H3O+]*[OH-] = 10^-14 [H3O+] = 10^-14/(1.64 x 10^-6) = 6.09 x 10^-9 Hope this helped Solution The F- will react with water to form HF and OH F- + H2O====> HF + OH- Kb for this reaction is 1.58 x 10^-11 At equilibrium, [F-] = .17 - x [HF] = x [OH-] = x The reactant is consumed by some amount x while the products are formed by some amount x Kb = concentration of products/reactants = x*x/(.17-x) = 1.58x 10^-11 x^2 + 1.58 x 10^-11x - 2.68 x 10^-12 = 0 Solve for x, x = 1.64 x 10^-6 [HF]= x = 1.64 x 10^-6 M [OH-] = x = 1.64 x 10^-6 M We know, [H3O+]*[OH-] = 10^-14 [H3O+] = 10^-14/(1.64 x 10^-6) = 6.09 x 10^-9 Hope this helped